Two identical beakers A and B contain equal volumes of two different… | JEE Main 2019 (8 Apr, Shift 1)

Q20JEE Main 2019 · 8 APR · SHIFT 1Medium

Newton's Law of Cooling

Two identical beakers A and B contain equal volumes of two different liquids at 60°C each and left to cool down. Liquid in A has density of 8 × 102 kg/m3 and specific heat of 2000 J kg–1 K–1 while liquid in B has density of 103 kg m–3 and specific heat of 4000 J kg–1 K–1. Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)

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Answer: (b)

Same volume $V$ , same area and emissivity, so the heat loss rate is the same for both at the same temperature.
Rate of fall $\frac{dT}{dt} = \frac{\text{heat loss rate}}{\rho V c} \propto \frac{1}{\rho c}$
A: $\rho c = 800 \times 2000 = 1.6\times10^6$ ; B: $\rho c = 1000 \times 4000 = 4\times10^6$
A has smaller $\rho c$ , so A cools faster and its curve lies below B's at all times.
So the answer is (b).

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